No, I am not familar with this theorem. I do not even understand the notation you are using.

However, I did think of an elementary way of proving

. We note that every element in

has to be expressed in terms of powers and linear combinations of

's. For example, consider

(here

) then

is basis for

and

is basis for

. Thus, basis for

is

. Therefore, any element can be written as

. And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the

-th roots. Let

be a basis for

where

's are products of the

's by above. If

then we can write

where

is a polynomial over

with

variables. Assume that

has the property that

. We wish to argue that

__all__ terms of

must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the

-th roots. Suppose this was not the case and there were two coefficients for

which were non-zero. Then in the expansion of

we would have a non-zero "cross term"

. But then this means

by the uniqueness of the representation of an element in terms of a basis.